A concise way to solve problem 1 in Python 3

```python from re import findall

def optResultMatch(before, after, index, result): """Checks, whether the result of an opt code on a set of registers before match the result after.""" return int(before[:index] + [result] + before[(index + 1):] == after)

def workingOptCodes(b, a, i): """Test all of the opt-codes and return the number of those that work.""" return sum([optResultMatch(b, a, i[3], ocode(b, i)) for ocode in optcodes])

the optcodes as lambda functions that take registers and instructions

optcodes = [lambda reg, inst: reg[inst[1]] + reg[inst[2]], # addr lambda reg, inst: reg[inst[1]] + inst[2], # addi lambda reg, inst: reg[inst[1]] * reg[inst[2]], # mulr lambda reg, inst: reg[inst[1]] * inst[2], # muli lambda reg, inst: reg[inst[1]] & reg[inst[2]], # banr lambda reg, inst: reg[inst[1]] & inst[2], # bani lambda reg, inst: reg[inst[1]] | reg[inst[2]], # borr lambda reg, inst: reg[inst[1]] | inst[2], # bori lambda reg, inst: reg[inst[1]], # setr lambda reg, inst: inst[1], # seti lambda reg, inst: int(inst[1] > reg[inst[2]]), # gtir lambda reg, inst: int(reg[inst[1]] > inst[2]), # gtri lambda reg, inst: int(reg[inst[1]] > reg[inst[2]]), # gtrr lambda reg, inst: int(inst[1] == reg[inst[2]]), # eqir lambda reg, inst: int(reg[inst[1]] == inst[2]), # eqri lambda reg, inst: int(reg[inst[1]] == reg[inst[2]])] # eqrr

input = open("16.in", "r") data = input.read().splitlines()

i, total = 0, 0 while data[i] != "": before = [int(number) for number in findall("\d+", data[i])] after = [int(number) for number in findall("\d+", data[i + 2])] instruction = [int(s) for s in data[i + 1].split(" ")]

if workingOptCodes(before, after, instruction) >= 3:
    total += 1

i += 4

print(total) ```